CHEMICAL EQUILIBRIUM ERT 207 ANALYTICAL CHEMISTRY SEMESTER 1, ACADEMIC SESSION 2015/16.

CHEMICAL EQUILIBRIUM

ERT 207 ANALYTICAL CHEMISTRY

SEMESTER 1, ACADEMIC SESSION 2015/16

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OVERVIEW

THE CHEMICAL COMPOSITION OF AQUEOUS SOLUTIONS

CHEMICAL EQUILIBRIUM THE EQUILIBRIUM STATE EQUILIBRIUM-CONSTANT

EXPRESSIONS TYPES OF EQUILIBRIUM CONSTANTS

IN ANALYTICAL CHEMISTRY APPLYING THE ION-PRODUCT

CONSTANT FOR WATER bblee@unimap

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OVERVIEW

SOLUBILITY-PRODUCT CONSTANTS

LE CHATELIER’S PRINCIPLE

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Electrolytes: It forms ions when dissolved in water (or

certain other solvents) and thus produce solutions that conduct electricity.

Strong electrolytes ionize essentially completely in a solvent, but weak electrolytes ionize only partially. These characteristics mean that a solution

of a weak electrolyte will only conduct electricity as well as a solution containing an equal concentration of a strong electrolytes.

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A salt is produced in the reaction of an acid with a base. Examples include NaCl, Na2SO4,

NaOOCH3

Table 1: Classification of Electrolytes

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Acids and Bases: According to Bronsted-Lowry concept, an acid donates protons and a base accepts protons.

An acid donates protons only in the presence of a proton acceptor (a base). Likewise, a base accepts protons only in the presence of a proton donor (an acid).

A conjugate base is formed when an acid loses a proton.

Acetate ion is the conjugate base of acetic acid.

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acid1 ↔ base1 + proton We refer to acid1 and base1 as a conjugate

pair. Every base accepts a proton to produce a

conjugate acid. base2 + proton ↔ acid2

Neutralization reaction: acid1 + base2 ↔ base1 + acid2

Many solvents are proton donors or proton acceptors & can induce basic or acidic behavior in solutes dissolved in them.

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In an aqueous solution of ammonia, water can donate a proton and acts as an acid with respect to the solute NH3:

Ammonia (base1) reacts with water (acid2), to give the conjugate base (base2) of the acid water.

Water acts as a proton acceptor, or base, in an aqueous solution of nitrous acid.

OHNHOHNH 423 Base1 Acid 2 Conjugate Conjugate Acid1 Base2

2322 NOOHHNOOH Base1 Acid 2 Conjugate Conjugate Acid1 Base2

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The conjugate base of the acid HNO2 is nitrate ion.

The conjugate acid of water is the hydrated proton, H3O+ (hydronium ion).

Figure 1: Possible

structures for the

hydronium ion.

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Amphiprotic species: Species that have both acidic and

basic properties. Example: Dihydrogen phopshate

OHPOHOHPOH 243342

Base1 Acid2 Acid1 Base2

OHHPOOHPOH 22442

Acid1 Base2 Base1 Acid2

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The simple amino acids are an important class of amphiprotic compounds that contain both a weak acid and a weak base functional group.

When dissolved in water, an amino acid, such as glycine, undergoes a kind of internal acid/base reaction to ptoduce a zwitterion. A zwitterion is an ion that has both a

positive and a negative charge.

COOCHNHCOOHCHNH 2322

Glycine zwitterion

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Water is the classic example of an amphiprotic solvent, that is, a solvent that can act either as an acid or as a base, depending on the solute.

Other common amphiprotic solvents are methanol, ethanol, and anhydrous acetic acid.

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Strengths of acids and bases Figure 2 shows the dissociation reactions

of a few common acids in water.

Figure 2: Dissociation reactions & relative strengths of some common acids & their conjugate base.

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The common strong bases include NaOH, KOH, Ba(OH)2, and the quaternary ammonium hydroxide R4NOH (R = an alkyl group: CH3 or C2H5).

The common strong acids include HCl, HBr, HI, HClO4, HNO3, the first proton in H2SO4, and the organic sulfonic acid RSO3H.

Perchloric and hydrochloric acids are strong acids in water.

If anhydrous acid, a weaker proton acceptor that water, is substituted as the solvent, neither of these acids undergoes complete dissociation.

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Equilibria below is established:

In a differentiating solvent, various acids dissociate to different degrees and have different strengths.

In a leveling solvent, several acids are completely dissociated and show the same strength.

22343 ClOCOOHCHHClOCOOHCHBase1 Acid 2 Acid 2 Base 2

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Many reactions used in analytical chemistry never result in complete conversion of reactants to products. Instead, they proceed to a state of chemical equilibrium in which the ratio of concentrations of reactants and products is constant.

Equilibrium-constant expressions are algebraic equations that describe the concentration relationship among reactants and products at equilibrium.

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The Equilibrium State: Consider the chemical reaction:

The rate of this reaction and the extent to which it proceeds to the right by monitoring the appearance of the orange-red color of the triiodide ion I3

-. The final position of a chemical

equilibrium is independent of the route to the equilibrium state.

OHIHAsOHIAsOH 23343 23

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The equilibrium state is altered by applying stress (temperature, pressure, concentration of reactant or product) to the system. These effects can be predicted

qualitatively from the Le Chatelier’s principle.

The Le Chatelier’s principle: It states that the position of an

equilibrium always shifts in such a direction as to relieve a stress that is applied to the system.

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An increase in temperature of a system alters the concentration relationship in the direction that tends to absorb heat, and an increase in pressure favors those participants that occupy a smaller total volume.

The mass-action effect is a shift in the position of an equilibrium caused by adding one of the reactants or products to a system.

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Equilibrium is a dynamic process. Although chemical reactions appear to

stop at equilibrium, in fact, the amounts of reactants and products are constant because the rates of the forward and reverse processes are exactly the same.

Chemical thermodynamics is a branch of chemistry that concerns the flow of heat & energy in chemical reactions. The position of a chemical equilibrium is

related to these energy changes.

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Equilibrium-constant expressions The influence of concentration or pressure on the position of a chemical equilibrium is conveniently described in quantitative terms by means of an equilibrium-constant expression.

They allow us to predict the direction and completeness of chemical reactions.

However, yields no information concerning the rate of reaction.

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Consider a generalized equation for a chemical equilibrium: wW + xX ↔ yY + zZ

The equilibrium-constant expression:

If a reactant or product is a pure liquid, a pure solid, or the solvent present in excess, no term for this species appears in the equilibrium-constant expression.

xw

zy

XW

ZYK

Molar concentration or partial pressure in

atmospheres if they are gas phase reactants or products.

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A temperature dependent equilibrium constant:

Table 1 summarizes the types of chemical equilibria and equilibrium constant that are of importance in analytical chemistry.

xw

y

XW

YK

xX

wW

zZ

yY

aa

aaK The activities of

species Y, Z, W, X

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Table 1: Equilibria & Equilibrium constants important in analytical chemistry

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Applying the ion-product constant for water

Aqueous solutions contain small concentrations of hydronium and hydroxide ions as a consquence of the dissolution reaction.

An equilibrium constant for this reaction can be formulated:

OHOHOH 322

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3

]OH[

]OH][OH[K

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When compared with the concentration of hydrogen and hydroxide ions:

At 25oC, Kw = 1.008 x 10-14. Table 2: Variation of KW with

temperature

]OH][OH[K]OH[K W 3

22

Ion-product constant for water

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Example 1: Calculate the hydronium and

hydroxide ion concentrations of pure water at 25oC and 100oC.

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Using solubility-product constants: Sparingly soluble salts are essentially

completely dissociated in saturated aqueous solution.

When an excess of barium iodate is equilibrated with water, the dissociation process is adequately described by: aqIOaqBasIOBa 3

223 2

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]sIOBa[

]IO][Ba[K

23

23

2

23

223 ]IO][Ba[K]s)IO(Ba[K sp

Solubility-product constant or Solubility

product

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Example 2: What mass (in grams) of Ba(IO3)2

(487 g/mol) can be dissolved in 500 ml of water at 25oC?

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The effect of a common ion in the solubility of a precipitate:

The common-ion effect is a mass-action effect predicted from Le Chatelier principle. The solubility of an ionic precipitate decreases when a soluble compound containing one of the ions of the precipitate is added to the solution.

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Example 3: Calculate the molar solubility of

Ba(NO3)2 in a solution that is 0.0200 M in Ba(NO3)2.

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When a change is imposed on a system at equilibrium the position of the equilibrium shifts in a direction that tends to reduce the effect of that change.

Henri Le Chatelier 1850-1936 Studied mining engineering. Interested in glass and ceramics.

It is like the “undo” button on your computer!

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Factors that Affect Equilibrium:i. Concentration ii. Temperature iii. Pressure (For gaseous systems only!)iv. The presence of a catalyst

(i) Effect of a Change in Concentration

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The reaction quotient for an equilibrium system is the same as the equilibrium expression, but the concentrations are NOT at equilibrium!

• Add more reactant Shift to products• Remove reactants Shift to reactants

• Add more product Shift to reactants

• Remove products Shift to products

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Changes in concentration are best understood in terms of what would happen to “Q” if the concentrations were changed.If Q< K then there are too many reactants, the reaction will shift in the forward direction (the products).

If Q>K then there are too many products, the reaction will shift to the reactants.

N2O4(g) 2NO2(g)Q = [NO2]2

[N2O4]

Keq at equilibri

um

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(ii) Effect of a Change in Pressure (Volume)

The system is initially at equilibrium.

The piston is pushed in, decreasing the volume and increasing the pressure. The system shifts in the direction that consumes CO2 molecules, lowering the pressure again.

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Decreasing volume:

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It only affects equilibrium systems with unequal moles of gaseous reactants and products.

Increase Pressure Stress of pressure is reduced by reducing

the number of gas molecules in the container.

There are 4 molecules of reactants vs. 2 molecules of products.

Thus, the reaction shifts to the product ammonia.

N2(g) + 3H2(g) = 2NH3(g)

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Increasing the volume: The system shifts in the direction that

increases its pressure. Decrease Pressure:

Stress of decreased pressure is reduced by increasing the number of gas molecules in the container.

There are two product gas molecules vs. one reactant gas molecule.

Thus, the reaction shifts to the products.

PCl5(g) = PCl3(g) + Cl2(g)

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(iii) Effect of a Change in Temperature The value of K changes with temperature

(We can use this to predict the direction of this change). Exothermic reaction – produces heat

(heat is a product)Adding energy shifts the equilibrium to

the left (away from the heat term). Endothermic reaction – absorbs energy

(heat is a reactant) Adding energy shifts the equilibrium to the right (away from the heat term).

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The meaning of K:K>1 the equilibrium position is far to

the right K<1 the equilibrium position is far to

the left

(iv) Presence of a Catalyst A catalyst lowers the activation energy and

increases the reaction rate. It will lower the forward and reverse

reaction rates.

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Therefore, a catalyst has NO EFFECT on a system at equilibrium! It just gets you to equilibrium faster!

Presence of an Inert Substance An inert substance is a substance that is

not- reactive with any species in the equilibrium system. These will not affect the equilibrium

system. If the substance does react with a

species at equilibrium, then there will be a shift!

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Example 4 Given:

S8(g) + 12O2(g) 8 SO3(g) + 808 kcals

What will happen when: Oxygen gas is added?

The reaction vessel is cooled?

Shifts to products

Shifts to Products – to replace heat

? ? ?

? ? ?

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The size of the container is increased?

Sulfur trioxide is removed?

A catalyst is added to make it faster?

V increases, Pressure decreases, shifts to more particles – to reactants !Shift to products to replace it !

No change !

? ? ?

? ? ?

? ? ?

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EXAMPLE 1Since OH- and H3O+ are formed only

from the dissolution of water, their concentration must be equal:

Then,

At 25C,

= 1.00 x 10-7 M

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]OH[]H[ 03WK]OH[]OH[ 22

3

WK]OH[]OH[ 3

143 10001 x.]OH[]OH[

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EXAMPLE 1At 100C,

= 7.0 x 10-7 M

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143 1049 x]OH[]OH[

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EXAMPLE 2The solubility-product constant for

Ba(IO3)2 is 1.57 x 10-9 (Appendix 2).Equilibrium between the solid and its

ions in solution is described:

The equilibrium reveals that 1 mol of Ba2+ is formed for each mole of Ba(IO3)2 that dissolves.

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23 2IOBa)s(IOBa

923

2 10571 x.IOBaK sp

EXAMPLE 2Molar solubility of Ba(IO3)2 = [Ba2+]

Since two moles of iodate are produced for each mole of barium ion, the iodate concentration is twice the barium ion concentration:

The equilibrium constant expression:

= 1.57 x 10-9bblee@unimap

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23 2 BaIO

32222 42 BaBaBa

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EXAMPLE 2

= 7.32 x 10-4 M

Since 1 mol of Ba2+ is produced for every mole of Ba(IO3)2,

Solubility = 7.32 x 10-4 MThe number of milimoles of Ba(IO3)2

dissolved in 500 ml of solution,

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2

4

10571/

x.Ba

mlx

mL

IOBammolx.IOBammol.No 50010327 234

23

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EXAMPLE 2The mass of Ba(IO3)2 in 500 ml is given

by: Mass Ba(IO3)2 =

= 0.178 g

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2323

4

)Ba(IO487050010327

mmol

IOBag.xIOBammolxx.

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EXAMPLE 3Molar solubility of Ba(IO3)2 = 0.5 [IO3

-]There are two sources of barium ions:

Ba(NO3)2 and Ba(IO3)2.The contribution from the nitrate is

0.0200 M, and that from the iodate is equal to the molar solubility, or 0.5 [IO3

-].

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2

102000 IO.Ba

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EXAMPLE 3The solubility-product expression,

Assumption:

Simplified equation:

=2.80 x 10-4 M bblee@unimap

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33 105712

102000

x.IOIO.

M.IO.Ba 020002

102000 3

2

92

3 1057102000 x.IO.

893 108570200010571 x..x.IO

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EXAMPLE 3Solubility of Ba(IO3)2 =0.5 [IO3

-] =0.5 x 2.80 x 10-4

= 1.40 x 10-4 M

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CHEMICAL EQUILIBRIUM ERT 207 ANALYTICAL CHEMISTRY SEMESTER 1, ACADEMIC SESSION 2015/16.

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Transcript of CHEMICAL EQUILIBRIUM ERT 207 ANALYTICAL CHEMISTRY SEMESTER 1, ACADEMIC SESSION 2015/16.